If `f(x)` is a function of x such that `(1)/((1+x)(1+x^(2)))=(A)/(1+x)+(f(x))/((1+x^(2))` for all `x in R` then `f(x)` is

Options

A. `(1-x)/(2)`
B. `(1+x)/(2)`
C. `1-x`
D. `1+x`

Answer

To solve the equation

\[ \frac{1}{(1+x)(1+x^2)} = \frac{A}{1+x} + \frac{f(x)}{1+x^2} \]

for the function \( f(x) \), we will follow these steps:

### Step 1: Assume the form of \( f(x) \)
Since \( f(x) \) is a function that needs to be expressed in partial fractions, we can assume it has the form:

\[ f(x) = bx + c \]

where \( b \) and \( c \) are constants.

### Step 2: Rewrite the equation
Substituting \( f(x) \) into the equation gives:

\[ \frac{1}{(1+x)(1+x^2)} = \frac{A}{1+x} + \frac{bx + c}{1+x^2} \]

### Step 3: Find a common denominator
To combine the right-hand side, we need a common denominator, which is \( (1+x)(1+x^2) \):

\[ \frac{A(1+x^2) + (bx + c)(1+x)}{(1+x)(1+x^2)} = \frac{A + Ax^2 + bx + bx^2 + c + cx}{(1+x)(1+x^2)} \]

### Step 4: Set the numerators equal
Now we equate the numerators from both sides:

\[ 1 = A + Ax^2 + bx + bx^2 + c + cx \]

### Step 5: Rearrange the equation
Rearranging gives us:

\[ 1 = (A + b)x^2 + (b + c)x + (A + c) \]

### Step 6: Compare coefficients
Since the left-hand side has no \( x^2 \) or \( x \) terms, we can set the coefficients equal to zero:

1. Coefficient of \( x^2 \): \( A + b = 0 \)
2. Coefficient of \( x \): \( b + c = 0 \)
3. Constant term: \( A + c = 1 \)

### Step 7: Solve the system of equations
From \( A + b = 0 \), we have:

\[ b = -A \]

From \( b + c = 0 \), substituting \( b \):

\[ -A + c = 0 \implies c = A \]

From \( A + c = 1 \):

\[ A + A = 1 \implies 2A = 1 \implies A = \frac{1}{2} \]

Now substituting \( A \) back to find \( b \) and \( c \):

\[ b = -\frac{1}{2}, \quad c = \frac{1}{2} \]

### Step 8: Write \( f(x) \)
Now substituting \( b \) and \( c \) into the expression for \( f(x) \):

\[ f(x) = -\frac{1}{2}x + \frac{1}{2} = \frac{1}{2} - \frac{1}{2}x \]

### Final Answer
Thus,

\[ f(x) = \frac{1}{2} - \frac{1}{2}x \]